_ ^ x ^2 x ;dx = - Vidyadip

MCQ

$\int_{}^{} {x{{\sec }^2}x\;dx} = $

A
$\tan x + \log \cos x + c$
B
$\frac{{{x^2}}}{2}{\sec ^2}x + \log \cos x + c$
C
$x\tan x + \log \sec x + c$
✓
$x\tan x + \log \cos x + c$

✓

Answer

Correct option: D.

$x\tan x + \log \cos x + c$

d
(d)$\int_{}^{} {x{{\sec }^2}x\,dx = x\tan x} - \int_{}^{} {\tan x\,dx} $
$ = x\tan x + \log (\cos x) + c.$

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